BITSAT Physics Test - 5 - JEE MCQ

When the sphere is half immersed, the system is in equilibrium.When the sphere is given a small push and displaced by 15 cm, the upthrust and the spring force will increase.
Therefore, spring force,

Increase buoyancy = Weight of displaced volume of liquid

Net restoring force,

Acceleration,

Frequency of oscillation,

This is the required solution.

Velocity of light remains unaffected by the motion of the observer. It is in fact apparent frequency/wavelength that changes. Velocity of light remains as such, i.e. c.
Hence, this is the required solution.

Due to insertion of glass plate, the entire fringe pattern shifts due to refraction of light; the fringe width does not change.

In the first container:

In the second container:

The ratio of the mass of the gas in the first container to that in the second container is given to be 2 : 3.
Therefore,  This is the required solution.

The value of the electric field for a solid charged sphere can be given as:
For internal points of sphere

For external points of sphere and on the surface

The value of the electric field for an infinitely long line charge can be given as:

The value of the electric field for a charged spherical shell can be given as:

Therefore, it is the correct option.

Acceleration of the bob will be g' = g cos θ
Therefore, the time period of the pendulum is given by

This is the required solution.

The angular half width  of the central maxima is given by sin θ λ/a
sin θ  = θ  radian when θ is small.


Total angular width of central maxima =2θ = 2 × 0.18 = 0.36°
Hence, this is the required solution.

In the processes, P₁ is isothermal and P₂ is adiabatic.

Also, form the graph,
W_1→2 = positive
As volume is increasing
Also, W_{2 →3} = negative
As volume is decreasing
Total work done = Area under graph
Therefore, W_{1 → 2} < W_{2 → 3}
Net work done, W = W_{1 → 2} + W_{2 → 3}
Therefore,W < 0 (negative)
Hence, it is the correct option.

The charge is hanging from the ceiling. After attaining the equilibrium position, the forces act on the charge q will be its weight, the electrostatic force and the tension in the string.

From the figure above,
Apply Lami's theorem,

We know,

This is the required solution.

When a p−n junction is formed, n-side have holes in the depletion region and p-side have electrons in the depletion region. In the absence of any external potential difference, drifting of minority charge carriers takes place in the depletion region leading to drift current and diffusion of majority charge carriers takes place outside the depletion region leading to diffusion current. Here, the drift current and depletion current are equal in magnitude and cancel out the effects of each other in the absence of any external potential difference. Hence, there will not be any steady current in the circuit.

From the relation of equivalence of charge carrier in extrinsic and intrinsic carrier, we have n_en_h = n_i² , where n_e, n_h & n_i represents electron density, hole density and number of intrinsic charge carrier density.
On substituting all the corresponding given values, we have

Number density of atoms in silicon specimen
= 5 × 10²⁸ atoms-m⁻³ = 5 × 10²² atoms cm⁻³.
Since, 1 atom of indium is doped in 5 × 10⁷ silicon atoms, so total number of indium atoms doped per cm³ of silicon will be

Given above is the schematic diagram of amplifier in common emitter configuration. From the diagram we can see that, in common emitter amplifier, the input signal is applied across base- emitter junction.

The voltage gain of an amplifier is initially constant and then decreases with the frequency of the signal.

Pure Cu is already an excellent conductor since it has a partially filled conduction band, furthermore, Cu forms a metallic crystal as opposed to the covalent crystals of silicon or germanium, so, the scheme of using an impurity to donate or accept an electron does not work for copper.
In fact, adding impurities to copper decreases the conductivity because an impurity tends to scatter electrons, impeding the flow of the current.

The output gate circuit will be as shown below.

= output is 1 when inputs are different, otherwise output is zero.

The PN junction will conduct only when it is forward biased i.e. when - 5 V is fed to it. So, it will conduct only for third-quarter part of the signal shown and when it conducts potential drop 5 V will be across both the resistors, so output voltage across R₂ is 2.5 V.
∴ V₀ = −2.5V
When input voltage is 5 V, PN junction is reversed biased. So, output voltage will be zero at that time.

The Boolean expression of this arrangement is:
For OR gate, 
Y₁ = A + B
For AND gate,
Y = Y₁⋅C = (A+B)⋅C
Using the truth table of AND gate, this expression is TRUE only when C is TRUE.

At point 'B', m will be moving horizontally
Let its speed be v₁
Let speed of M be v₂
Applying linear momentum conservation in horizontal, as there is no force on system in horizontal, we will get,

As the semicircular track is smooth,
Applying energy conservation we will get,

As shown a block of mass M is lying over rough horizontal surface. Let μ be the coefficient of kinetic friction between the two surfaces in contact. The force Of friction between the block and horizontal surface is given by
F = μR = μMg , (∵ R = Mg)
To move the block without acceleration, the force (P)required will be just equal to the force of friction, ie,
P = F = μR
If d is the distance moved, then work done is given by
W = P × d = μRd

m : mass per unit length
∴  rate of mass leaving the hose per sec

The initial kinetic energy of the gun is KE_I = 0 and potential energy  and work = 0.
The final kinetic energy is KE_F = 0 and potential energy, PE_F = Mg(4S)
Energy is conserved,

Total length of wire is 'l' and mass is m so each side length = ℓ/4 and mass = m/4.
Work done by gravity is calculate by the displacement of COM in vertical direction